Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume th…

Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume th…

394. 字符串解码 394. Decode String 题目描述 给定一个经过编码的字符串,返回它解码后的字符串. 编码规则为: k[encoded_string],表示其中方括号内部的 encoded_string 正好重复 k 次.注意 k 保证为正整数. 你可以认为输入字符串总是有效的,输入字符串中没有额外的空格且输入的方括号总是符合格式要求的. 此外,你可以认为原始数据不包含数字,所有的数字只表示重复的次数 k,例如不会出现像 3a 或 2[4] 的输入. 每日一算法2019/6/1…

Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_stringinside the square brackets is being repeated exactly k times. Note that kis guaranteed to be a positive integer. You may assume that…

[题目] Total Accepted: 10087 Total Submissions: 25510 Difficulty: Medium Contributors: Admin Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repe…

题目 题目 s = "3[a]2[bc]", return "aaabcbc". s = "3[a2[c]]", return "accaccacc". s = "2[abc]3[cd]ef", return "abcabccdcdcdef". 大概意思是根据指定的格式将字符串进行展开 思路 递归的思路 递归计算出括号最里面的字符串,依次再处理外面一层的字符串,每个单元内的字符串类似于一…

Given an encoded string, return it's decoded string.The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.You may assume that…

Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume th…

［抄题］: Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may ass…

2018-11-14 17:56:12 问题描述: 问题求解: 方法一.递归求解 最直观的解法就是递归来求了,并且很显然的这个问题可以使用递归来进行求解. public String decodeString(String s) { StringBuffer sb = new StringBuffer(); int num = 0; for (int i = 0; i < s.length(); i++) { if (Character.isDigit(s.charAt(i))) { num =…